Perimeter, Area, Volume

Perimeter, Area, and Volume of Standard Shapes and Solids

Perimeter Formulas

Shape	Formula
Rectangle	\(P = 2(l + w)\)
Square	\(P = 4s\)
Circle (circumference)	\(C = 2\pi r = \pi d\)
Triangle	\(P = a + b + c\)

Area Formulas

Shape	Formula
Rectangle	\(A = lw\)
Square	\(A = s^2\)
Triangle	\(A = \tfrac{1}{2}bh\)
Parallelogram	\(A = bh\)
Trapezium	\(A = \tfrac{1}{2}(a+b)h\)
Circle	\(A = \pi r^2\)
Sector	\(A = \tfrac{\theta}{360}\pi r^2\)

Volume Formulas

Solid	Formula
Rectangular prism	\(V = lwh\)
Cylinder	\(V = \pi r^2 h\)
Cone	\(V = \tfrac{1}{3}\pi r^2 h\)
Sphere	\(V = \tfrac{4}{3}\pi r^3\)
Pyramid	\(V = \tfrac{1}{3}Ah\) (where \(A\) = base area)

Surface Area Formulas

Solid	Formula
Rectangular prism	\(SA = 2(lw + lh + wh)\)
Cylinder	\(SA = 2\pi r^2 + 2\pi r h\)
Sphere	\(SA = 4\pi r^2\)
Cone	\(SA = \pi r^2 + \pi r l\) (where \(l\) = slant height)

Worked Example — Cylinder

A water tank is cylindrical with radius \(r = 0.6\) m and height \(h = 1.4\) m.

\[V = \pi r^2 h = \pi (0.6)^2 (1.4) = \pi \times 0.36 \times 1.4 = 0.504\pi \approx 1.583 \text{ m}^3\]

Capacity: \(1.583 \times 1000 = 1583 \text{ L}\).

\[SA = 2\pi(0.6)^2 + 2\pi(0.6)(1.4) = 2\pi(0.36 + 0.84) = 2\pi(1.2) \approx 7.54 \text{ m}^2\]

Worked Example — Composite Shape

A garden bed consists of a rectangle (5 m × 3 m) with a semicircle of diameter 3 m on one end.

\[A_{\text{rect}} = 5 \times 3 = 15 \text{ m}^2\]

\[A_{\text{semicircle}} = \tfrac{1}{2}\pi\left(\tfrac{3}{2}\right)^2 = \tfrac{1}{2}\pi(1.5)^2 = \tfrac{1}{2}\pi(2.25) \approx 3.53 \text{ m}^2\]

\[A_{\text{total}} \approx 15 + 3.53 = 18.53 \text{ m}^2\]

EXAM TIP: Always include units in every line of working. Examiners award method marks partly based on correct unit handling. State the formula, substitute values, then evaluate.

COMMON MISTAKE: Using the diameter instead of the radius in circle/cylinder/sphere formulas. Double-check: \(r = d/2\).

Perimeter, Area, Volume

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